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Re: How many positive integers divide 35^12 but not 35^11? [#permalink]

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13 Feb 2014, 01:03

As usual Brunnel has a good solution ready.However, I solved it in a different way.I'm metioning it here as this method will help identify those "uncommon" factors.

35^11 is not divisible by 35^12.Moreover, 35^11 has 7^11*5^11 as its factor, while 35^12 has 7^12*5^12 as its factor. (Highest powers of 7 &5 are 11 & 12 respectively)

Hence the remaining uncommon factors are 7^12*5^0, 7^12*5^1,....,7^12*5^11 (remember 7^12,5^12 has already been considered as 35^12) , i.e. 12and5^12*7^0,5^12*7^1,...,5^12*7^11 (remember 7^12,5^12 has already been considered as 35^12) , i.e. 12

Re: How many positive integers divide 35^12 but not 35^11? [#permalink]

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12 Apr 2015, 12:01

1

This post receivedKUDOS

The trick is to understand that they are looking for the number of factors for each of them . Factorize them and you will see that..

35^12 = 7^12 *5^12. The number of factors would be (12+1)(12+1) = 16935^11 = 7^11 *5^11. The number of factors would be (11+1)(11+1) = 144

The difference is 25.

Another problem I though about. Difference in the sum of the positive integers that divide 35^2 and 35^3. Note the powers are different to make the calculations possible. If you read the GMATclub math book, you will see the formula to calculate the sum of the of factors of an integer is, if N= (a^m)(b^n)Sum = (a^(m+1)-1)(b^(n+1)-1)/(a-1)(b-1)

You can factorize the calculations and finally get to workable numbers. I am almost certain that something like this will not come on GMAT.